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Chapter 9: Problem 1216

At \(100 \mathrm{~K}\) and \(0.1\) atmospheric pressure, the volume helium gas is 10 liters. If volume and pressure are doubled, its temperature will change to (A) \(127 \mathrm{~K}\) (B) \(400 \mathrm{~K}\) (C) \(25 \mathrm{~K}\) (D) \(200 \mathrm{~K}\)

Short Answer

Expert verified
The temperature of helium gas will change to \(200 \,\text{K}\) when the volume and pressure are doubled. The correct answer is (D) \(200 \,\text{K}\).

Step by step solution

01

Write down the Ideal Gas Law formula

We will use the Ideal Gas Law formula, which is represented by: \(PV = nRT\) Here, P = pressure, V = volume, n = amount of gas in moles, R = gas constant, and T = temperature in kelvin.
02

Analyze the given information

We are given the initial volume (\(V_1\)), pressure (\(P_1\)), and temperature (\(T_1\)) of the helium gas. We are also told that both the volume and pressure are doubled, which results in a change in the temperature (\(T_2\)). Therefore, we need to find the value of \(T_2\). Initial conditions: \(P_1=0.1\) atm, \(V_1=10\) L, \(T_1=100\) K. Final conditions: \(P_2=2P_1\), \(V_2=2V_1\), \(T_2=\) unknown.
03

Convert the given pressure to the same units

The units of pressure should be consistent within the formula. The Gas Constant (R) used in the given exercise is \(8.314 \,\text{J/molK}\), thus we need to convert the given pressure from atm to Pa (Pascals). The conversion factor is as follows: 1 atm = 101325 Pa. Now we can convert the given pressure: \(P_1 = 0.1 \,\text{atm} \times \frac{101325 \,\text{Pa}}{1 \,\text{atm}} = 10132.5 \,\text{Pa}\)
04

Apply the Ideal Gas Law formula for the initial condition

Now, let's use the Ideal Gas Law to find the unknown quantity of moles (n) in the helium gas. Rearranging the formula for calculating n, we get: \(n = \frac{P_1V_1}{RT_1}\). Substituting the given values: \(n = \frac{10132.5 \,\text{Pa} \times 10 \,\text{L}}{8.314 \,\text{J/molK} \times 100 \,\text{K}}\) Note: 1 L = 0.001 m³, we need to convert L to m³: \(n= \frac{10132.5 \,\text{Pa} \times 0.01 \,\text{m}^3}{8.314 \,\text{J/molK} \times 100 \,\text{K}}\)
05

Solve for n

Calculate the value of n: \(n \approx 0.122 \,\text{moles}\)
06

Calculate the final conditions and apply the Ideal Gas Law formula

We know that the pressure and volume are doubled, so we have the final conditions: \(P_2 = 2P_1 = 20265 \,\text{Pa}\) \(V_2 = 2V_1 = 0.02 \,\text{m}^3\) Now we apply the Ideal Gas Law formula again for the final conditions: \(P_2V_2 = nRT_2\) Rearranging the formula for calculating \(T_2\), we get: \(T_2 = \frac{P_2V_2}{nR}\). Substituting the values: \(T_2 = \frac{20265 \,\text{Pa} \times 0.02 \,\text{m}^3}{0.122 \,\text{moles} \times 8.314 \,\text{J/molK}}\)
07

Solve for T_2

Calculate the value of T_2: \(T_2 \approx 200 \,\text{K}\) The temperature of helium gas will change to 200 K when the volume and pressure are doubled. Therefore, the correct answer is (D) \(200 \,\text{K}\).

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Most popular questions from this chapter

If \(\mathrm{rms}\) speed of a gas is $\mathrm{v}_{\mathrm{rms}}=1840 \mathrm{~m} / \mathrm{s}\( and its density \)\rho=8.99 \times 10^{-2} \mathrm{~kg} / \mathrm{m}^{3}$, the pressure of the gas will be (A) \(1.01 \times 10^{3} \mathrm{Nm}^{-2}\) (B) \(1.01 \times 10^{5} \mathrm{Nm}^{-2}\) (C) \(1.01 \times 10^{7} \mathrm{Nm}^{-2}\) (D) \(1.01 \mathrm{Nm}^{-2}\)

1 mole of gas occupies a volume of \(100 \mathrm{~m} 1\) at \(50 \mathrm{~mm}\) pressure. What is the volume occupied by two moles of gas at $100 \mathrm{~mm}$ pressure and at same temperature (A) \(50 \mathrm{ml}\) (B) \(200 \mathrm{ml}\) (C) \(100 \mathrm{ml}\) (D) \(500 \mathrm{~m} 1\)

A gas at 1 atmosphere and having volume \(100 \mathrm{ml}\) is mixed with another gas of equal moles at \(0.5 \mathrm{~atm}\) and having \(50 \mathrm{ml}\) in flask of one litre, what is the final pressure? (A) \(0.125 \mathrm{~atm}\) (B) \(0.75 \mathrm{~atm}\) (C) 1 atm (D) \(0.5 \mathrm{~atm}\)

One mole of ideal monoatomic gas \([\gamma=(5 / 3)]\) is mixed with one mole of diatomic gas \([\gamma=(7 / 5)]\). What is \(\gamma\) for the nixture? \(\gamma\) denotes the ratio of specific heat at constant oressure to that at constant volume. (A) \((35 / 23)\) (B) \((23 / 15)\) (C) \((3 / 2)\) (D) \((4 / 3)\)

For a gas if ratio of specific heats at constant pressure and volume is \(\gamma\), then value of degrees of freedom is (A) \([2 /(\gamma-1)]\) (B) \([(25) / 2](\gamma-1)\) (C) \([(3 \gamma-1) /(2 \gamma-1)]\) (D) \((9 / 2)(\gamma-1)\)
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